Coulomb's law
Electric Fields - OCR A-Level Physics
Worked Example
Two point charges of +3.0 μC and -5.0 μC are separated by 0.20 m. Calculate (a) the force between them, and (b) the electric field strengthThe force per unit positive chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). at a point in an electric field. Measured in N C⁻¹ or V m⁻¹. at the midpoint.
Show Solution
1
Identify
$Q_{1} = 3.0 \times 10⁻⁶ C, Q_{2} = -5.0 \times 10⁻⁶ C, r = 0.20\;\text{m}.$
2
(a) Apply CoulombThe SI unit of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. One coulomb is the chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). transferred by a current of 1 A in 1 second.'s law
F = Q₁Q₂ / (4πε₀r²) = (8.99 × 10⁹)(3.0 × 10⁻⁶)(5.0 × 10⁻⁶) / (0.20)² = 0.135 / 0.04 = 3.4 N (attractive).
3
(b) At the midpoint, distance from each $charge = 0.10 m$.
4
E due to +3.0 μC = Q/(4πε₀r²) = (8.99 × 10⁹)(3.0 × 10⁻⁶)/(0.10)² = 2.70 × 10⁶ N C⁻¹ (pointing away from +Q, i.e. towards -Q).
5
E due to -5.0 μC = (8.99 × 10⁹)(5.0 × 10⁻⁶)/(0.10)² = 4.50 × 10⁶ N C⁻¹ (pointing towards -Q).
6
Both fields point in the same direction (from + towards -), so E_total = 2.70 × 10⁶ + 4.50 × 10⁶ = 7.2 × 10⁶ N C⁻¹.
Answer
F = 3.4 N (attractive); E at midpoint = 7.2 × 10⁶ N C⁻¹ directed from the positive charge towards the negative charge.